Electromagnetic Induction Ques 5

5. A circular disc of radius $ 0.2$ meter is placed in a uniform magnetic field of induction $\frac{1}{p}(Wb / m^{2})$ in such a way that its axis makes an angle of $60^{\circ}$ with $\vec{B}$. The magnetic flux linked with the disc is:

[2008]

(a) $0.02 $ $Wb$

(b) $0.06 $ $Wb$

(c) $0.08 $ $Wb$

(d) $0.01 $ $Wb$

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Answer:

Correct Answer: 5.(a)

Solution:

  1. (a) Here, $B=\frac{1}{p}(Wb / m^{2})$

$\theta=60^{\circ}$

Area normal to the plane of the disc

$=pr^{2} \cos 60^{\circ}=\frac{pr^{2}}{2}$

Flux $=B \times$ normal area

$=\frac{0.2^{\prime} 0.2}{2}=0.02 $ $Wb$